样例 1:
输入：["doaf","agai","dcan"]，["dog","dad","dgdg","can","again"]
输出：["again","can","dad","dog"]
解释：
d o a f
a g a i
d c a n
矩阵中查找，返回 ["again","can","dad","dog"]。
样例 2:
输入：["a"]，["b"]
输出：[]
解释：
a
矩阵中查找，返回 []。
题解
使用 HashMap 的版本。 将所有的单词的 Prefix 都加入 HashMap 中。HashMap 的 value 用户判断这是一个 prefix 还是一个 word。 如果是 prefix 就是 false，如果是 word 就是 true.
public class Solution {
public static int[] dx = {0, 1, -1, 0};
public static int[] dy = {1, 0, 0, -1};
/** * @param board: A list of lists of character * @param words: A list of string * @return: A list of string */
public List<String> wordSearchII(char[][] board, List<String> words) {
if (board == null || board.length == 0) {
return new ArrayList<>();
}
if (board[0] == null || board[0].length == 0) {
return new ArrayList<>();
}
boolean[][] visited = new boolean[board.length][board[0].length];
Map<String, Boolean> prefixIsWord = getPrefixSet(words);
Set<String> wordSet = new HashSet<>();
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
visited[i][j] = true;
dfs(board, visited, i, j, String.valueOf(board[i][j]), prefixIsWord, wordSet);
visited[i][j] = false;
}
}
return new ArrayList<String>(wordSet);
}
private Map<String, Boolean> getPrefixSet(List<String> words) {
Map<String, Boolean> prefixIsWord = new HashMap<>();
for (String word : words) {
for (int i = 0; i < word.length() - 1; i++) {
String prefix = word.substring(0, i + 1);
if (!prefixIsWord.containsKey(prefix)) {
prefixIsWord.put(prefix, false);
}
}
prefixIsWord.put(word, true);
}
return prefixIsWord;
}
private void dfs(char[][] board,
boolean[][] visited,
int x,
int y,
String word,
Map<String, Boolean> prefixIsWord,
Set<String> wordSet) {
if (!prefixIsWord.containsKey(word)) {
return;
}
if (prefixIsWord.get(word)) {
wordSet.add(word);
}
for (int i = 0; i < 4; i++) {
int adjX = x + dx[i];
int adjY = y + dy[i];
if (!inside(board, adjX, adjY) || visited[adjX][adjY]) {
continue;
}
visited[adjX][adjY] = true;
dfs(board, visited, adjX, adjY, word + board[adjX][adjY], prefixIsWord, wordSet);
visited[adjX][adjY] = false;
}
}
private boolean inside(char[][] board, int x, int y) {
return x >= 0 && x < board.length && y >= 0 && y < board[0].length;
}
}
更多题解参考：九章官网Solution